MCQ Questions for Class 11 Engineering Physics Mechanical Properties Of Solids Quiz 5

When a certain weight is suspended from a long uniform wire, its length increases by

$1\ cm$ . If the same weight is suspended from another wire of the same material and length but having a diameter half of the first one, the increase in length will be

0%
$0.5cm$
0%
$2cm$
0%
$4cm$
0%
$8cm$

A thick rope of density

$\rho$ and length

$L$ is hung from a rigid support. The increase in length of the rope due to its own weight is (

$Y$ is the Young's modulus)

0%
$\dfrac {0.1}{4Y} \rho L^{2}g$
0%
$\dfrac {1}{2Y} \rho L^{2}g$
0%
$\dfrac {\rho L^{2}g}{Y}$
0%
$\dfrac {\rho Lg}{Y}$

Explanation

Take a small section

$\delta y$ at distance

$y$ from top as shown in figure. Let area of rope be A

Stress due to this section =

$\rho A g \delta y / A = \rho d \delta y$

Strain on section y =

$\delta l / y$

Using Stress = Y \times Strain

$\rho g \delta y = Y \times \dfrac{\delta l} {y}$

$\Delta l = \dfrac {\rho g} {y} \int_{0}^{L} y \delta y$

$\Delta l = \dfrac {\rho g L^{2}} {2 Y}$

Answer.

$\dfrac {\rho g L^{2}} {2 Y}$

When the tension in a metal wire is

${T}_{1}$ , its length is

${l}_{1}$ . When the tension is

${T}_{2}$ , its length is

${l}_{2}$ . The natural length of wire is

0%
$\cfrac { { T }_{ 2 } }{ { T }_{ 1 } } ({ l }_{ 1 }+{ l }_{ 2 })$
0%
${ T }_{ 1 }{ l }_{ 1 }+{ i }_{ 2 }{ l }_{ 2 }$
0%
$\cfrac { { { l }_{ 1 }T }_{ 2 }-{ { l }_{ 2 }T }_{ 1 } }{ { T }_{ 2 }-{ T }_{ 1 } }$
0%
$\cfrac { { { l }_{ 1 }T }_{ 2 }+{ { l }_{ 2 }T }_{ 1 } }{ { T }_{ 2 }+{ T }_{ 1 } }$

Explanation

$\dfrac { \Delta { l } }{ { l } } =\dfrac { F }{ AY } \\ Thus,\quad \dfrac { { l }_{ 1 }-l }{ l } =\dfrac { { T }_{ 1 } }{ AY } \quad and\quad \dfrac { { l }_{ 2 }-l }{ l } =\dfrac { { T }_{ 2 } }{ AY }$
Solve the 2 equations to get,

$AY=\dfrac { { T }_{ 1 }l }{ { l }_{ 1 }-l } ,\quad giving\quad { T }_{ 2 }l({ l }_{ 1 }-l)={ T }_{ 1 }l({ l }_{ 2 }-l)\\ or\quad l=\dfrac { { T }_{ 2 }{ l }_{ 1 }-{ T }_{ 1 }{ l }_{ 2 } }{ { T }_{ 2 }-{ T }_{ 1 } }$

A wire stretched

$1mm$ by a force of

$1kN$ . How far would a wire of the same material and length but of four times that diameter be stretched by the same force?

0%
$\dfrac {1}{2} mm$
0%
$\dfrac {1}{4} mm$
0%
$\dfrac {1}{8} mm$
0%
$\dfrac {1}{16} mm$

Explanation

$\Delta l=\dfrac { FL }{ AY }$
Since

$F,L$ and

$Y$ are same in both cases, extension depends on Area, and is inversely proportional.

$4$ times diameter means

${4}^{2}$ times or 16 times Area. Thus, extension

$= 1/16$ times

$1 mm = 1/16 mm$

Bulk modulus of water is

$2\times 10^{9}N/m^{2}$ . The change in pressure required to increase the density of water by

$0.1\%$ is

0%
$2\times 10^{9}N/{m^{2}}$
0%
$2\times 10^{8}N/{m^{2}}$
0%
$2\times 10^{6}N/{m^{2}}$
0%
$2\times 10^{4}N/{m^{2}}$

Explanation

$B=-V\dfrac{dP}{dV}$

Thus,

$\Delta V=-\dfrac{\Delta pV}{B}$ , where p is change in pressure.

Thus,

$V'-V=-\dfrac{\Delta pV}{B}$ or,

$V'=V(1-\dfrac{\Delta p}{B})$

Now,

$\rho '=\dfrac{m}{V'}=\dfrac{m}{V(1-\dfrac{\Delta p}{B})}=\dfrac{\rho}{(1-\dfrac{\Delta p}{B})}$

Given

$\rho ' =1.001\rho$ , or

$\dfrac{1}{1.001}= 1-\dfrac{p}{2\times{10}^{9}}$

or,

$\Delta p=\dfrac{1}{1001}\times 2\times {10}^{9}$ , which is nearly

$2\times{10}^{6}$

Each of the pictures shows four objects tied together with rubber bands being pulled to the right across a horizontal frictionless surface by a horizontal force

$F$ . All the objects have the same mass; all the rubber bands obey Hooke's law and have the same equilibrium length and the same force constant. Which of these pictures is drawn most correctly?

A ball falling in a lake of depth

$200\ m$ shows a decrease of

$0.1\%$ in its volume at the bottom. The bulk modulus of the elasticity of the material of the ball is (take

$g = 10\ m/s^{2})$ .

0%
$10^{9}N/m^{2}$
0%
$2\times 10^{9}N/m^{2}$
0%
$3\times 10^{9}N/m^{2}$
0%
$4\times 10^{9}N/m^{2}$

Explanation

Bulk Modulus is given by;

$B=-V\dfrac{dP}{dV}$
Now, change in pressure,

$dP=\rho gh=1000 \times 10\times 200=2\times {10}^{6}$
Also, given

$\dfrac{dV}{V}=-0.001$
Thus,

$B=-\dfrac{2\times{10}^{6}}{-0.001}=2\times{10}^{9}$

Young's modulis of brass and steel are

$10\times 10^{10} N/m$ and

$2\times 10^{11} N/m^{2}$ , respectively. A brass wire and a steel wire of the same length are extended by

$1\ mm$ under the same force. The radii of the brass and steel wires are

$R_{B}$ and$ R_{S}$$, respectively. Then

0%
$R_{S} = \sqrt {2} R_{B}$
0%
$R_{S} = \dfrac {R_{B}}{\sqrt {2}}$
0%
$R_{S} = 4R_{B}$
0%
$R_{S} = \dfrac {R_{B}}{4}$

Explanation

$\delta l\quad =\quad \dfrac { FL }{ AE } \\ \Rightarrow \quad 1\quad =\quad \dfrac { { F }.L }{ { { A }_{ B } }.{ E }_{ B } } \quad =\quad \dfrac { { F }.L }{ { { A }_{ S } }.{ E }_{ S } } \quad or\quad { A }_{ B }.10\times { 10 }^{ 10 }\quad =\quad { A }_{ S }.2\times { 10 }^{ 11 }\\ Thus,\quad { A }_{ B }\quad =\quad 2A_{ S }\quad or\quad { R }_{ B }\quad =\quad { R }_{ S }\sqrt { 2 }$

$5\ kg$ rod of square cross section

$5\ cm$ on a side and

$1\ m$ long is pulled along a smooth horizontal surface by a force applied at one end. The rod has a constant acceleration of

$2\ m/s^{2}$ . Determine the elongation in the rod. (Young's modulus of the material of the rod is

$5\times 10^{3} N/m^{9})$ .

0%
Zero, as for elongation to be there, equal and opposite forces must act on the rod
0%
Non-zero but cannot be determine from the given situation
0%
$4\mu m$
0%
$16\mu m$

Two wires of the same material and same mass are stretched by the same force. Their length are in the ratio

$2:3$ . Their elongations are in the ratio

0%
$3:2$
0%
$2:3$
0%
$4:9$
0%
$9:4$

Explanation

Let the lengths of the two wires be

$2l$ and

$3l$ .
Two materials have same mass.
Thus we get

$\rho(2l)A_1=\rho(3l)A_2$
Thus

$2A_1=3A_2$
Now their elongations are given as

$\Delta l_1=\dfrac{F2l}{A_1Y}$ and

$\Delta l_2=\dfrac{F3l}{A_2Y}$
Thus we get

$\dfrac{\Delta l_1}{\Delta l_2}=\dfrac{4}{9}$

A rubber rope of length

$8\ m$ is hung from the ceiling of a room. What is the increase in length of rope due to its own weight? (Given : Young's modulus of elasticity of rubber

$= 5\times 106\ N/m$ and density of rubber

$= 1.5\times 10^{3} kg/ m^{3}$ . Take

$g = 10\ m/s^{2})$ .

0%
$1.5mm$
0%
$6mm$
0%
$24mm$
0%
$96mm$

Explanation

Weight of rubber

$W=mg=\rho ALg=$

$8\times 1.5\times 10^3\times A\times g$ , where

$A$ is cross sectional area.

Elongation due to own weight is half of the elongation due to point load.

$\Delta L=\dfrac{WL}{2AY}=\dfrac{8\times 1.5\times 10^3\times A\times L\times g}{2A\times 5\times 10^6 }=96\ mm$

The valve

$V$ in the bent tube is initially kept closed. Two soap bubbles

$A$ (smaller) and

$B$ (larger) are formed at the two open ends of the tube.

$V$ is now opened and air can flow freely between the bubbles.

0%
There will be no change in the size of the bubbles
0%
The bubbles will become of equal size
0%
$A$ will become smaller and $B$ will become larger
0%
The sizes of $A$ and $B$ will be interchanged

Two wires

$A$ and

$B$ have the same cross section and are made of the same material, but the length of wire

$A$ is twice that of

$B$ . Then, for a given load.

0%
The extension of $A$ will be twice that of $B$
0%
The extensions of $A$ and $B$ will be equal
0%
The strain in $A$ will be half that in $B$
0%
The strains in $A$ and $B$ will be equal

Explanation

We know that

$Y = \dfrac {FL}{Al}$

$\therefore l = \dfrac {FL}{AY}$
Since, the two wires are made of the same material, Young's modulus

$Y$ is same for both. Since, load

$F$ and the cross-sectional area

$A$ are the same,

$\therefore l \propto L$
i.e., extension is proportional to the length of the wire.
Hence, option (a) is correct. The strain in a wire is given by

$\dfrac {l}{L} = \dfrac {4F}{AY}$
Since

$F, A$ and

$Y$ are the same, the strains in wires

$A$ and

$B$ will be equal. Thus, the correct options are (a) and (d).

Identical springs of steel and sopper

$(Y_s > Y_{cu})$ are equally stretched. Then

0%
less work is done on steel spring
0%
less work is done on copper spring
0%
equal work is done on both the springs
0%
data not complete

Explanation

$Work=Force\times Displacement$
Now, Displacement is given as:

$\Delta l=\dfrac{FL}{AY}$ or,

$F=\dfrac{AY\Delta l}{L}$
Thus,

$Work=\dfrac{AY{(\Delta l)}^{2}}{L}$
Since, similar wires are used and extension is same, thus, A,

$\Delta l$ and L are same. Also,

${Y}_{s}>{Y}_{Cu}$ .
Thus, work on steel is greater than work on copper.

Which of the following affects the elasticity of a substance?

0%
Hammering and annealing
0%
Change in temperature
0%
Impurity in substance
0%
All of the above

A steel wire of length

$4.7\ m$ and cross-sectional area

$3\times 10^{-6} m^{2}$ stretches by the same amount as a copper wire of length

$3.5\ m$ ad cross-sectional area of

$4\times 10^{-6} m^{2}$ under a given load. The ratio of Young's modulus of steel to that of copper is

0%
$1.8$
0%
$3.6$
0%
$0.6$
0%
$8.7$

Explanation

We know,

$\Delta l =\dfrac{FL}{AY}$

Given, load and extension are same, thus,

$\dfrac{{Y}_{1}}{{Y}_{2}}=\dfrac{\dfrac{{L}_{1}}{{A}_{1}}}{\dfrac{{L}_{2}}{{A}_{2}}}=\dfrac{\dfrac{4.7}{3}}{\dfrac{3.5}{4}}=1.8$

A wire is stretched by a force

$F$ . If

$s$ is the strain developed and

$Y$ is Young's modulus of material of wire, then work done per unit volume is

0%
$\displaystyle\frac{Ys^2}{2}$
0%
$\displaystyle\frac{s^2}{2Y}$
0%
$\displaystyle\frac{1}{2}Fs$
0%
$\displaystyle\frac{Y}{2s^2}$

Explanation

Work done per unit volume is same as energy per unit volume which is given by:

$\dfrac{1}{2}\times Stress \times Strain=1/2\times (Y\times Strain)\times Strain=\dfrac{1}{2}\times Y\times {(Strain)}^{2}=\dfrac{Y{s}^{2}}{2}$

A iron bar of length

$l$ cm and cross section

$A$

$cm^2$ is pulled by a force of

$F$ dynes iron ends so as to produce an elongation

$\Delta l$ cm. Which of the following statement is correct?

0%
Elongation is inversely proportional to length
0%
Elongation is directly proportional to cross section $A$
0%
Elongation is inversely proportional to $A$
0%
Elongation is directly proportional to Young's module

Explanation

$\Delta l=\dfrac { FL }{ AY }$
Thus, extension is inversely proportional to area. Only C statement is correct

The longitudinal extension of any elastic material is very small. In order to have an appreciable change, the material must be in the form of

0%
thin block of any cross section
0%
thick block of any cross section
0%
long thin wire
0%
short thin wire

Explanation

$\Delta l =\dfrac{FL}{AY}$
Thus, extension is directly proportional to length and inversely to area. Thus for appreciable extension, length must be long and area less, or long and thin wire

A uniform plank is resting over a smooth horizontal floor and is pulled by applying a horizontal force at its one end. Which of the following statements are not correct?

0%
Stress developed in plank material is maximum at the end at which force is applied and decrease linearly to zero at the other end
0%
A uniform tensile stress is developed in the plank material
0%
Since plank is pulled at one end only, plank starts to accelerate along direction of the force. Hence, no stress is developed in the plank material
0%
None of the above

Explanation

Since, force is applied at one end only, the plank starts to accelerate along the direction of this force and a stress is developed in its material. Hence, option (c) is incorrect.
To calculate stress in the material of the plank, at a distance

$x$ from the end at which force is applied, the free-body diagrams are considered as shown in the figure.

$\therefore F' = \dfrac {m}{l} (l - x) a ....(i)$

$F - F' = \dfrac {m}{l} xa$
From these two equations,

$F' = \dfrac {F(l - x)}{l}$

$\therefore Stress = \dfrac {F'}{A} = \dfrac {F(l - x)}{Al}$
Where

$A$ is cross-sectional area of the plank.
It shows that stress varies linearly with

$x$ . It is maximum at

$x = 0$ and zero at

$x = l$ .
Hence, option (a) is correct.
1557064_157944_ans_19f7b13c39a742b8a7462755bf50d8ae.png

A spherical ball contracts in volume by

$0.02\%$ when subjected to a pressure of

$100$ atmosphere. Assuming one atmosphere

$=10^8\ Nm^{-2}$ , the bulk modulus of the material of the ball is

0%
$0.02\times 10^5\ N/m^2$
0%
$0.02\times 10^7\ N/m^2$
0%
$50\times 10^7\ N/m^2$
0%
$50\times 10^9\ N/m^2$

Explanation

$B=-V\dfrac{dP}{dV}$
Given

$dP=100atm, \dfrac{dV}{V}=-0.0002$
Thus,

$B=-\dfrac{100}{-0.0002}=5\times{10}^{5}$ in atm units.

$1\ atm = {10}^{5}Pa$ , thus in SI units,

$B=5\times {10}^{10}=50\times{10}^{9}\ N/{m}^{2}$

The bulk modulus of a perfectly rigid body is

0%
infinity
0%
zero
0%
some finite value
0%
non-zero constant

Explanation

$B=-V\dfrac{dP}{dV}$
Perfectly rigid body means there can be no change in volume, or

$dV$ tends to zero. Thus, B tends to infinity.

A metal rod of Young's modulus

$2\times 10^{10}Nm^{-2}$ undergoes an elastic strain of

$0.06\%$ . The energy per unit volume stored in

$Jm^{-3}$ is

0%
$3600$
0%
$7200$
0%
$10800$
0%
$14400$

Explanation

Energy per unit volume

$=\dfrac{1}{2}\times Stress \times Strain=\dfrac{1}{2} \times (Strain \times Y)\times Strain=\dfrac{1}{2} \times Y \times {(Strain)}^{2}$
Thus, we get,

$0.5 \times 2\times {10}^{10} \times {(0.0006)}^{2}=3600$

There are two wires of same material and same length while the diameter of second wire is two times the diameter of first wire, then the ratio of extension produced in the wires by applying same load will be

0%
$1:1$
0%
$2:1$
0%
$1:2$
0%
$4:1$

Uniform rod of mass

$m$ , length

$l$ , area of cross-section

$A$ has Young's modulus

$Y$ . If it is hanged vertically, elongation under its own weight will be :

0%
$\displaystyle\frac{mgl}{2AY}$
0%
$\displaystyle\frac{2mgl}{AY}$
0%
$\displaystyle\frac{mgl}{AY}$
0%
$\displaystyle\frac{mgY}{Al}$

Explanation

$Y=\displaystyle\frac{Fl}{A\Delta l}\Rightarrow \Delta l=\frac{Fl}{YA}=\frac{mgl}{YA}$ (in this question whole mass is considered on centre of mass and hence for calculating max displacement, we are calculating he displacement of centre of mass only)

A steel ring of radius

$r$ and cross sectional area

$A$ is fitted onto a wooden disc of radius

$R(R>r)$ . If the Young's modulus of steel is

$Y$ , then the force with which the steel ring is expanded is

0%
$AY(R/r)$
0%
$AY(R-r)/r$
0%
$(Y/A)[(R-r)/r]$
0%
$Yr/AR$

Explanation

Let

$T$ be the tension in the ring, then

$Y=\displaystyle\frac{T.2\pi r}{A.2\pi(R-r)}=\frac{Tr}{A(R-r)}$

$\therefore T=\displaystyle\frac{YA(R-r)}{r}$

Two wires of same material and length but cross-sections in the ratio

$1:2$ are used to suspend the same loads. The extensions in them will be in the ratio

0%
$1:2$
0%
$2:1$
0%
$4:1$
0%
$1:4$

Explanation

Let the cross section of wire be A , 2A

$\Delta L_1=\dfrac{Fl}{AY}$

$\Delta L_2=\dfrac{Fl}{2AY}$

so,

$\dfrac{\Delta L_1}{\Delta _2}=2/1$

The length of an iron wire is

$L$ and area of cross-section is

$A$ . The increase in length is

$l$ on applying the force

$F$ on its two ends. Which of the statement is correct?

0%
Increase in length is inversely proportional to its length
0%
Increase in length is proportional to area of cross-section
0%
Increase in length is inversely proportional to area of cross-section
0%
Increase in length is proportional to Young's modulus.

Explanation

$l=\displaystyle\frac{FL}{YA}\Rightarrow l\propto \frac{1}{A}$

From a steel wire of density

$\rho$ is suspended a brass block of density

$\rho_b$ . The extension of steel wire comes to

$e$ . If the brass block is now fully immersed in a liquid of density

$\rho_{\gamma}$ the extension becomes

$e'$ . The ratio

$\dfrac{e}{e^{'}}$ will be

0%
$\displaystyle\dfrac{\rho_b}{\rho_b-\rho_l}$
0%
$\displaystyle\dfrac{\rho_b-\rho_l}{\rho_b}$
0%
$\displaystyle\dfrac{\rho_b-\rho}{\rho_l-\rho}$
0%
$\displaystyle\dfrac{\rho_l}{\rho_b-\rho_l}$

Explanation

Since everything else

$(L, A, Y)$ is kept constant, extension will be directly proportional to the Force, which is mg or

${\rho}_{b} Vg$
After immersing, tension will reduce due to buoyant force and will become

${\rho}_{b} Vg-{\rho}_{\gamma}Vg$
Thus,

$\dfrac{e}{e'}=\dfrac{{\rho}_{b} Vg}{{\rho}_{b} Vg-{\rho}_{\gamma}Vg}=\dfrac{{\rho}_{b}}{{\rho}_{b}-{\rho}_{\gamma}}$

A cube at temperature

$0^oC$ is compressed equally from all sides by an external pressure

$P$ . By what amount should its temperature be raised to bring it back to the size it had before the external pressure was applied. The bulk modulus of the material of the cube is

$B$ and the coefficient of linear expansion is

$\alpha$ .

0%
$P/B\alpha$
0%
$P/3B\alpha$
0%
$3\pi \alpha/ B$
0%
$3B/P$

Explanation

Bulk modulus

$B=\displaystyle\frac{-P}{(\Delta V/V)}=\frac{-PV}{\Delta V}$ .......(1)
and

$\Delta V= \gamma V\Delta T=3\alpha V. T$ or

$\displaystyle\frac{-V}{\Delta V}=\frac{1}{3\alpha. T.}$ .....(2)
From eqs.

$(1)$ and

$(2)$ ,

$B=P/(3\alpha . T)$ or

$\displaystyle T=\frac{P}{3\alpha B}$

$A$ and

$B$ are two wires. The radius of

$A$ is twice that of

$B$ . They are stretched by the same load. Then the stress on

$B$ is

0%
equal to that on $A$
0%
four times that on $A$
0%
two times that on $A$
0%
half that on $A$

Explanation

$stress=\dfrac{Force}{Area}$

$\therefore \displaystyle stress \propto \dfrac{1}{\pi r^2}$
Thus,

$\dfrac{S_B}{S_A}=\left[\dfrac{r_A}{r_B}\right]^2=(2)^2\Rightarrow S_B=4S_A$

When a pressure of

$100$ atmosphere is applied on a spherical ball, then its volume reduces to

$0.01\%$ . The bulk modulus of the material of the rubber in

$dyne/cm^2$ is :

0%
$10\times 10^{12}$
0%
$100\times 10^{12}$
0%
$1\times 10^{12}$
0%
$1000\times 10^{12}$

Explanation

$B=-V\dfrac{dP}{dV}=\dfrac{100}{0.01/100}=10^6\ atm$
which is

$10^{11}\ N/{m}^{2}$
Now,

$1N={10}^{5}\ dynes$ and

$1{m}^{2}={10}^{4}\ {cm}^{2}$
Thus, we get,

$B={10}^{12}\ dyne/{cm}^{2}$

Two wire

$A$ and

$B$ are of the same material. Their lengths are in the ratio of

$1:2$ and the diameter are in the ratio

$2:1$ . If they are pulled by the same force, then increase in length will be in the ratio of

0%
$2:1$
0%
$1:4$
0%
$1:8$
0%
$8:1$

Explanation

We know that Young's modulus

$Y=\displaystyle\dfrac{F}{\pi r^2}\times \dfrac{L}{l}$

Since

$Y$ ,

$F$ are same for both the wires, we have,

$\displaystyle\dfrac{1}{r^2_1}\dfrac{L_1}{l_1}=\dfrac{1}{r^2_2}\dfrac{L_2}{l_2}$

or,

$\displaystyle\dfrac{l_1}{l_2}=\dfrac{r^2_2\times L_1}{r^2_1\times L_2}=\displaystyle\dfrac{(D_2/2)^2\times L_1}{(D_1/2)^2\times L_2}$

or,

$\displaystyle\dfrac{l_1}{l_2}=\displaystyle\dfrac{D^2_2\times L_1}{D^2_1\times L_2}=\displaystyle\dfrac{D^2_2}{(2D_2)^2}\times \dfrac{L_2}{2L_2}=\dfrac{1}{8}$

so,

$l_1:l_2=1:8$

If a rubber ball is taken at the depth of

$200m$ in a pool, its voulme decreases by

$0.1\%$ . If the density of the water is

$1\times 10^3kg/m^3$ and

$g=10m/s^2$ , then the volume elasticity in

$N/m^2$ will be

0%
$10^8$
0%
$2\times 10^8$
0%
$10^9$
0%
$2\times 10^9$

Explanation

$K=\displaystyle\frac{\Delta P}{\Delta V/V}=\frac{hpg}{\Delta V/V}=\displaystyle\frac{200\times 10^3\times 10}{0.1\times 100}=2\times 10^9$

Two wires are made of the same material and have the same volume. However wire

$1$ has cross-sectional area

$A$ and wire

$2$ has cross-sectional area

$3A$ . If the length of wire

$1$ increases by

$\Delta x$ on applying force

$F$ , how much force is needed to stretch wire

$2$ by the same amount?

0%
$4F$
0%
$6F$
0%
$9F$
0%
$F$

Explanation

$\textbf{Step 1 - Drawing the both figure. }$

Since volume is same

$\because A \times l_{1} = 3A \times l_{2} \Rightarrow l_{2} = \dfrac {l_{1}}{3}$

$\textbf{Step 2 - Writing the Young modules for both figure}$

Since Young module

$Y = \dfrac {Stress}{Strain} = \dfrac {\dfrac {F}{A}}{\dfrac {\triangle x}{l}}$

As material is same so

$Y$ will be same

$\dfrac {\dfrac {F}{A}}{\dfrac {\triangle x}{l_{1}}} = \dfrac {\dfrac {F'}{3A}}{\dfrac {\triangle x}{\dfrac {l_{1}}{3}}}$

$\Rightarrow F' = 9F$

Hence (C) is correct.

A mass m is suspended from a wire. Change in length of the wire is

$\Delta l$ . Now the same wire is stretched to double its length and the same mass is suspended from the wire. The change in length in this case will become (it is assumed that elongation in the wire is within the proportional limit)

0%
$\Delta l$
0%
$2\Delta l$
0%
$4\Delta l$
0%
$8\Delta l$

Explanation

Elongation is given by

$\Delta l=\dfrac {Fl}{AY}=\dfrac {Fl}{\frac {V}{l}Y}=\dfrac {Fl^2}{VY}$
SInce the force, volume and Y are constant

$\Delta l\propto l^2$
New length is double the original length, so the elongation would be

$\dfrac {\Delta l_2}{\Delta l_1}=\left (\dfrac {l_2}{l_1}\right )^2$

$\Delta l_2=\Delta l_1\left (\dfrac {2l}{l}\right )^2=\Delta l_1(4)$
Option C.

The maximum load that a wire can sustain is

$W$ . If the wire is cut to half its value, the maximum load itcan sustain is

0%
$W$
0%
$\dfrac{w}{2}$
0%
$\dfrac{w}{4}$
0%
$2W$

Explanation

stress

$= \dfrac{F}{A}$

$F = stress \times A$

still the value of maximum force or load remains the same as it dod'nt depend on length of wire.

A uniform steel rod of cross-sectional area

$A$ and length

$L$ is suspended so that it hangs vertically. The stress at the middle point of the rod is :

0%
$\dfrac{1}{2}\rho gL$
0%
$\dfrac{1}{4}\rho gL$
0%
$\rho gL$
0%
None of these

Explanation

Force acting at the mid point of the rod is

$T=\dfrac {M}{2}(g)=\dfrac {\rho SL}{2}(g)$

$\therefore \sigma =\dfrac {T}{S}=\dfrac {1}{2}\rho gl$
Option A.

Identify the case when an elastic metal rod does not undergo elongation:

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it is pulled with a constant acceleration on a smooth horizontal surface
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it is pulled with constant velocity on a rough horizontal surface
0%
it is allowed to fall freely
0%
all of the above

The bulk modulus of water is

$2.0 \times 10^{9} N/m^{2}$ The pressure required to increase the density of water by

$0.1\%$ is :

0%
$2.0\times 10^{3} N/m^{2}$
0%
$2.0\times 10^{6} N/m^{2}$
0%
$2.0\times 10^{5} N/m^{2}$
0%
$2.0\times 10^{7} N/m^{2}$

Explanation

$\rho^{'} = \rho(1+\dfrac{dp}{B})$

change in density

$\Delta \rho =\rho^{'} -\rho = \dfrac{\rho dp}{B}$

$\dfrac{\Delta \rho}{\rho} = \dfrac{dp}{B}$

$\dfrac{\Delta \rho}{\rho} \times 100 = 0.1$

$\dfrac{\Delta \rho}{\rho} = 0.001$

$\Delta P = B\dfrac{\Delta \rho}{\rho} = 2 \times 10^9 \times 0.001 = 2\times 10^6 N/m^2$

Vessel of

$1 \times 10^{-3} m^{3}$ volume contains an oi. If a pressure of

$1.2 \times 10^{5} N/m^{2}$ is applied on it, thenvolume decreases by

$0.3\times 10^{-3}m^{3}$ . The bulk modulus of oil is

0%
$6\times 10^{10}N/m^{2}$
0%
$4\times 10^{5}N/m^{2}$
0%
$2\times 10^{7}N/m^{2}$
0%
$1\times 10^{6}N/m^{2}$

Explanation

Bulk modulus of the oil is given by

$B=-\dfrac {\Delta P}{\Delta V/V}=-\dfrac {(1.2\times 10^5)(1\times 10^{-3})}{-0.3\times 10^{-3}}=4\times 10^5N/m^2$
Option B.

Two wires of the same material have lengths in the ratio 1:2 and their radii are in the ratio

$1:\sqrt{2}$ . If they are stretched by applying equal forces, the increase in their lengths will be in the ratio :

0%
$2$
0%
$\sqrt{2}:2$
0%
$1:1$
0%
$1:2$

Explanation

$\displaystyle \Delta\ L =\frac{Fl}{YA}; \ \frac{\Delta l}{\Delta l_{2}}=\frac{l_{1}}{l_{2}}\times \left ( \frac{r_{2}}{r_{1}} \right )^{2}=\frac{1}{2}\times \left ( \frac{\sqrt{2}}{1} \right )^{2}=1:1$

The area of cross-section of a wire of length

$1.1$ meter is

$1\:mm^{2}$ . It is loaded with 1 kg. If Young's modulus of copper is

$1.1\times 10^{11}N/m^{2}$ , then the increase in length will be (If

$g=10\:m/s^{2}$ ) :

0%
$0.01\ mm$
0%
$0.075 \ mm$
0%
$0.1\ mm$
0%
$0.15\ mm$

Explanation

$\displaystyle y=\dfrac{F/A}{\dfrac{\Delta l}{l}}$

$\displaystyle \Delta l=\frac{Fl}{Ay}$

$\displaystyle \Delta l=\frac{Mgl}{Ay}$

$\displaystyle \Delta l=\frac{1\times 10\times 1.1}{1\times 10^{-6}\times 1.1\times 10^{11}}=1\times 10^{-4}=0.1\:mm$

How much force is required to produce an increase of 0.2% in the length of a brass wire of diameter 0.6 mm? [Young modulus for brass

$=0.9\times 10^{11}N/m^{2}$ ]:-

0%
Nearly 17 N
0%
Nearly 34 N
0%
Nearly 51 N
0%
Nearly 68 N

Explanation

$\displaystyle Y=\frac{F/A}{\frac{\Delta l}{l}}$ $\displaystyle F=\frac{YA\Delta l}{l}$

$\displaystyle F=Y.\pi r^{2}\frac{\Delta l}{l}$

$\displaystyle F=0.9\times 10^{11}\times 3.14\times \left ( \frac{0.6}{2}\times 10^{-3} \right )^{2}\times \frac{0.2}{100}$

$\displaystyle F=0.9\times 10^{11}\times 3.14\times \times 0.09\times 10^{-6}\times 2\times 10^{-3}$

$\displaystyle F=51\:N$

The Young's modulus of a rubber string

$8\ cm$ long and density

$1.5\ kg/m^ {3}$ is

$5\times 10^{8}N/m^{2},$ is suspending on the ceiling in a room. The increase in the length due to its own weight will be:

0%
$9.6\times 10^{-5}\ m$
0%
$9.6\times 10^{-11}\ m$
0%
$9.6\times 10^{-3}\ m$
0%
$9.6\ m$

Explanation

$\displaystyle Y=\dfrac{F/A}{\dfrac{\Delta l}{l}}$

$\displaystyle F=\dfrac{Mg}{2}$

$\displaystyle \Delta l=\dfrac{FA}{AY}$

$\displaystyle F=\frac{Al\rho g}{2}$

$\displaystyle \Delta l=\dfrac{Al\rho gl}{2AY}=\frac{\rho l^{2}g}{2y}$

$\displaystyle \Delta l=\dfrac{1.5\times 64\times 10^{-14}\times 10}{2\times 5\times 10^{8}}$

$\Delta l =9.6 \times 10^{-11}m$

On increasing the length by

$0.5 mm$ in a steel wire of length

$2 m$ and area of cross-section

$2\:cm^{2}$ , the force required is [

$Y$ for steel

$=2.2\times 10^{11}N/m^{2}$ ]:

0%
$1.1\times 10^{5}\:N$
0%
$1.1\times 10^{4}\:N$
0%
$1.1\times 10^{3}\:N$
0%
$1.1\times 10^{2}\:N$

Explanation

Force,

$F = \dfrac{YA \Delta l}{L} = \dfrac{2.2 \times 10^{11} \times 2 \times 10^{-4} \times 0.5 \times 10^{-3}}{2} = 1.1 \times 10^3 N$

A force of

$10^{3}$ N stretches the length of a hanging wire by

$1$ millimeter. The force required to stretch a wire of same material and length but having four times the diameter by

$1$ millimeter is :

0%
$4\times 10^{3}$ N
0%
$16\times 10^{3}$ N
0%
$\displaystyle \frac{1}{4}\times 10^{3}$ N
0%
$\displaystyle \frac{1}{16}\times 10^{3}$ N

Explanation

$\displaystyle Y=\frac{F/A}{\Delta l/l}$ ; $Y$ & $\displaystyle \frac{\Delta l}{l}=constant$

$\displaystyle F\propto A\propto r^{2}$ $\displaystyle \frac{F_{2}}{F_{1}}=\left ( \frac{r_{2}}{r_{1}} \right )^{2}=\left ( \frac{4r}{r} \right )^{2}=16$

$F_{2}=16\times 10^{3}\:N$

A fixed volume of iron is drawn into a wire of length

$l$ . The extension produced in this wire by a constant force

$F$ is proportional to :

0%
$\displaystyle \dfrac{1}{l^{2}}$
0%
$\displaystyle \dfrac{1}{l}$
0%
$l^{2}$
0%
$l$

Explanation

let the volume be

$V$ and area

$A$

$Y = \dfrac{Fl}{A \Delta l}$

$\Delta l = \dfrac{Fl}{YA}$

$V = A \times l$

$A =\dfrac{V}{l}$

$\Delta l = \dfrac{Fl^2}{YV}$

The diameter of a brass rod is 4 mm and Young's modulus of brass is

$9\times 10^{10}N/m^{2}$ . The force required to stretch by 0.1% of its length is :-

0%
$360\pi \:N$
0%
36 N
0%
$144\pi \times 10^{3}\:N$
0%
$36\pi \times 10^{5}\:N$

Explanation

$Y = \dfrac{FL}{A \Delta l}$

$F = \dfrac{Y \Delta l A}{L} = \dfrac{YA\times 0.1L}{L} = 0.1YA = 0.1 \times 9 \times 10^{10}\times \pi \times (\dfrac{4\times 10^{-3}}{2})^2 = 360\pi$

The dimensions of two wires

$A$ and

$B$ are the same. But their materials are different, Their load- extension graphs are shown. If

$Y_{A}$ and

$Y_{B}$ are the values of Young's modulus of elasticity of

$A$ and

$B$ respectively then :

0%
$Y_{A} >Y_{B}$
0%
$Y_{A}< Y_{B}$
0%
$Y_{A} =Y_{B}$
0%
$Y_{B} =2Y_{A}$

Explanation

$Y = \dfrac{FL}{\Delta l A}$

$\Delta l = \dfrac{FL}{YA}$

For same load elongation in A is greater than in B therefore

$\Delta l_A > \Delta l_B$

$\dfrac{FL}{Y_A A} > \dfrac{FL}{Y_B A}$

$Y_B >Y_A$

CBSE Questions for Class 11 Engineering Physics Mechanical Properties Of Solids Quiz 5 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

CBSE Questions for Class 11 Engineering Physics Mechanical Properties Of Solids Quiz 5 - MCQExams.com

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Practice Class 11 Engineering Physics Quiz Questions and Answers

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